Position wave function of two electrons

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SUMMARY

The discussion focuses on calculating the position wave function of a system of two free electrons characterized by momenta k1 and k2. The user attempts to derive Psi_(k1,k2)(x1,x2) using the momentum eigenfunctions and the delta function normalization = (2Pi)^3 Delta(k-k'). Key insights include the realization that using definite momentum values leads to non-localized wave functions, necessitating the use of wave packets to achieve localization in position space. The importance of adhering to the Pauli exclusion principle for identical fermions is also emphasized.

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  • Quantum mechanics fundamentals, particularly wave functions and eigenstates
  • Understanding of momentum operators and their eigenvectors
  • Familiarity with Fourier transforms in quantum mechanics
  • Knowledge of the Pauli exclusion principle and its implications for fermions
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Students and researchers in quantum mechanics, particularly those studying multi-particle systems and wave function analysis. This discussion is beneficial for anyone looking to deepen their understanding of electron behavior and wave function calculations in quantum physics.

Faust90
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Hi,

I want to calculate the position-wave-function of a system of two free electrons with momenta k1 and k2 (vectors).

1. Homework Statement


So, I want to have Psi_(k1,k2)(x1,x2) for a state |k1,k2>

I also know that <k'|k> = (2Pi)^3 Delta(k-k')

The Attempt at a Solution



I tried the following:

Psi(x1,x2)=<x1,x2|k1,k2>=<x1,x2|1|k1,k2>=Integral<x1,x2|k1',k2'><k1',k2'|k1,k2>dk1dk2

The first term <x1,x2|k1',k2'> are the momenta eigenfunction in Space presentation, the second term is the given delta-function:

= Integral Exp(ik1'x1)Exp(ik2'x2) Delta(k1-k1')Delta(k2-k2')
=Exp(ik1x1)Exp(ik2x2)

Is that right?
I'm a bit confused about k1,k2 and k1',k2'Best regards :-)
Faust
 
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Faust90 said:
Psi(x1,x2)=<x1,x2|k1,k2>=<x1,x2|1|k1,k2>=Integral<x1,x2|k1',k2'><k1',k2'|k1,k2>dk1dk2

The first term <x1,x2|k1',k2'> are the momenta eigenfunction in Space presentation, the second term is the given delta-function:

= Integral Exp(ik1'x1)Exp(ik2'x2) Delta(k1-k1')Delta(k2-k2')
=Exp(ik1x1)Exp(ik2x2)
Aren't you running in circles here? If you can replace <x1,x2|k1',k2'> by Exp(ik1'x1)Exp(ik2'x2), why can't you do that with <x1,x2|k1,k2>?

If you take definite values of ##k##, then the wave function is not localized: the uncertainty on position is infinite. You need to take a more physically acceptable starting point, such as a wave packet centered around ##k_1## and ##k_2##. You can then Fourier transform it to get the wave function in position space.

Remember also that the total wave function has to obey the Pauli principle.
 
Hi,

thanks for your answer :-)

I'm still a bit confused. Let me summary again what I have:

Two electrons characterized by a wave vector k, where I know that the normalization is :<k|k'>=(2Pi)^3 Delta(k-k').
Now I have the state |k1,k2> and I shall construct the space wave function.

My first questions:1. The |k>,|k'> are the eigenvectors of the momentum operator, or?
2. Does the state |k1,k2> mean that I have definite values of k?
 
Faust90 said:
My first questions:

1. The |k>,|k'> are the eigenvectors of the momentum operator, or?
2. Does the state |k1,k2> mean that I have definite values of k?
Yes to both.
 
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